Parallel axes theorem - Statement :
The moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its centre of gravity and the product of the mass of the body and the square of the distance between the two axes.

*Theorems of moment of inertia*

*(i) Parallel axes theorem*

*Statement*

*The
moment of inertia of a body about any axis is equal to the sum of its moment of
inertia about a parallel axis through its centre of gravity and the product of
the mass of the body and the square of the distance between the two axes.*

*Proof*

Let us consider a body having its centre of
gravity at G as shown in Fig.. The axis XX′ passes through the centre of gravity and is
perpendicular to the plane of the body. The axis X_{1}X_{1}′ passes through the point O
and is parallel to the axis XX′ . The distance between the two parallel axes
is *x*.

Let the body be divided into large number of
particles each of mass *m *. For a
particle* P *at a distance* r *from O, its moment of inertia about
the* *axis X_{1}OX_{1}′ is equal to *m r* * ^{2}*.

The moment of inertia of the whole body about
the axis X_{1}X_{1}′ is given by,

*I = **Σ** mr ^{2 } ???(1)*

From the point P, drop a perpendicular PA to the
extended OG and join PG.

In the ∆*OPA*,

*OP ^{2} = OA^{2} + AP ^{2}*

*r ^{2} = x^{2} + 2xh + h^{2}
+ AP^{2 } ???(2)*

But from ∆ *GPA*,

*GP *^{2}* *=* GA*^{2}* *+*
AP *^{2}

*y ^{2} = h ^{2} *+

Substituting equation (3) in
(2),

*r ^{2} = x ^{2} + 2xh + y ^{2 } *..(4)

Substituting equation (4) in (1),

*I _{0}
= *

*= **Σ**mx ^{2} + *

*= Mx ^{2} + My^{2} + 2x*

Here *My ^{2}*

*Σ **(mg) (h) = 0** *(or)* Σ **mh** *= 0 [since* **g** *is a constant]

equation (5) becomes, *I _{0}= Mx^{2} + I_{G} *

Thus the parallel axes theorem is proved.

*(ii) Perpendicular axes theorem*

*Statement*

*The moment of inertia of a plane laminar body
about an axis perpendicular to the plane is equal to the sum of the moments of
inertia about two mutually perpendicular axes in the plane of the lamina such
that the three mutually perpendicular axes have a common point of intersection.*

*Proof*

Consider a plane lamina
having the axes *OX* and *OY* in the plane of the lamina as shown
Fig. The axis *OZ *passes through* O *and is* *perpendicular to the plane of the lamina. Let the lamina be
divided into a large number of particles, each of mass *m*. A particle at *P* at a
distance *r* from *O* has coordinates (x,y).

∴*r ^{2} = x^{2}+y^{2 }*

The moment of inertia of the
particle *P* about the axis *OZ* = *m
r ^{2}*. The moment of inertia of the whole lamina about the axis

*I _{Z}= *

The moment
of inertia of the whole lamina about the axis *OX* is

*I _{x} =*

Similarly,
*I _{y}=* Σ

From eqn.
(2), *I =* Σ *mr ^{2}*

*I =*
Σ*mx ^{2}+*

* I _{z}
= I_{x}+ I_{y}*

which proves the
perpendicular axes theorem.

**List : Moment of Inertia of
different bodies**

**Body : Axis
of Rotation**

Thin Uniform Rod

Axis passing through its
centre of
gravity and perpendicular to its length

Axis passing through the end
and perpendicular to its length.

Thin Circular Ring

Axis passing through its centre
and perpendicular to its plane.

Axis passing through its Diameter

Axis passing through a Tangent

Circular Disc

Axis passing through
its centre and perpendicular to its plane.

Axis passing through its diameter

Axis passing through a tangent

Solid Sphere

Axis passing through its diameter

Axis passing through a tangent

Solid Cylinder

Its own axis

Axis passing through its centre
and perpedicular to its length

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